How to use 'or' operator in regex properly?

Question

I am trying Javascript's regular expression.
I understand that ' | ' is used to or-ing two regular expression.
I created a regex /^a*|b*$/ , and I want it to detect any string that contains only charater of 'a' or 'b'.
But when I try /^a*|b*$/.test('c') , it produces true ?
What I am missing understading of ' | ' operator?

Here's my code:

let reg = /^a*|b*$/;
< undefined
reg.test('c');
< true

Answer 1

| has very low precedence. ^a*|b*$ matches

• either ^a*
• or b*$

ie either a string beginning with 0 or more 'a's or a string ending with 0 or more 'b's. (Because matching 0 'a's is allowed by the regex, any string will match (because every string has a beginning).)

To properly anchor the match on both sides, you need

/^(?:a*|b*)$/

(the (?: ) construct is a non-capturing group).

You could also use

/^a*$|^b*$/

instead.

Note that both of these regexes will only match strings like aa , bbbbbb , etc., but not aba . If you want to allow the use of mixed a/b characters in a string, you need something like

/^(?:a|b)*$/

Answer 2

The OR in your example will split the expression in these alternatives: ^a* and b*$ .

You can use groups to delimit the alternatives

Something like

/^(a*|b*)$/

This will match empty strings, strings that contains only a characters and strings that contain only b characters.

If you're looking to match strings that contain both a and b characters, you can use something like:

/^[ab]*$/